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Date: Mon, 21 Sep 2015 19:46:11 +0200
From: Oleg Nesterov <oleg@redhat.com>
To: Peter Zijlstra <peterz@infradead.org>
Cc: Boqun Feng <boqun.feng@gmail.com>,
        "Paul E. McKenney" <paulmck@linux.vnet.ibm.com>,
        linux-kernel@vger.kernel.org, linux-doc@vger.kernel.org,
        Jonathan Corbet <corbet@lwn.net>, Michal Hocko <mhocko@kernel.org>,
        David Howells <dhowells@redhat.com>,
        Linus Torvalds <torvalds@linux-foundation.org>,
        Will Deacon <will.deacon@arm.com>
Subject: Re: [PATCH] Documentation: Remove misleading examples of the
	barriers in wake_*()
Message-ID: <20150921174611.GA28059@redhat.com>
References: <1441674841-11498-1-git-send-email-boqun.feng@gmail.com> <20150909192822.GM4029@linux.vnet.ibm.com> <20150910021612.GC18494@fixme-laptop.cn.ibm.com> <20150910175557.GA20640@redhat.com> <20150917130125.GL3816@twins.programming.kicks-ass.net> <20150917170111.GA29215@redhat.com> <20150918064956.GQ3816@twins.programming.kicks-ass.net>
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On 09/18, Peter Zijlstra wrote:
>
> the text is correct, right?

Yes, it looks good to me and helpful.

But damn. I forgot why exactly try_to_wake_up() needs rmb() after
->on_cpu check... It looks reasonable in any case, but I do not
see any strong reason immediately.

Say,

	p->sched_contributes_to_load = !!task_contributes_to_load(p);
	p->state = TASK_WAKING;

we can actually do this before "while (p->on_cpu)", afaics. However
we must not do this before the previous p->on_rq check.

So perhaps this rmb() helps to ensure task_contributes_to_load() can't
happen before p->on_rq check...

As for "p->state = TASK_WAKING" we have the control dependency in both
cases. But the modern fashion suggests to use _CTRL(). Although cpu_relax()
should imply barrier(), but afaik this is not documented.

In short, I got lost ;) Now I don't even understand why we do not need
another rmb() between p->on_rq and p->on_cpu. Suppose a thread T does

	set_current_state(...);
	schedule();

it can be preempted in between, after that we have "on_rq && !on_cpu".
Then it gets CPU again and calls schedule() which clears on_rq.

What guarantees that if ttwu() sees on_rq == 0 cleared by schedule()
then it can _not_ still see the old value of on_cpu == 0?

Oleg.